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| Question 903107:  In a math book I am using / reading the author states that the expression n + (n-3) simplifies to n(n-1). But he does not show steps for this or give an explanation.
 I checked the expression on many web-based math sites (including your site) and none show the simplification of the original expression as n(n-1).
 One site states that n + (n-3) simplifies to the following expression (with steps shown):
 ------------------------------------------------
 Simplifying
 n + n(n + -3)
 Reorder the terms:
 n + n(-3 + n)
 n + (-3 * n + n * n)
 n + (-3n + n2)
 Combine like terms: n + -3n = -2n
 -2n + n2
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 Can you state whether the expression n + n(n-3) can simplify to n(n-1) and if so, can you please show the steps.
 Thank you in advance.
 Greg
 Answer by AlgebraLady88(44)
      (Show Source): 
You can put this solution on YOUR website! Ok, let's take a look at the equation n + n (n- 3)
 For the part  n (n +3) , we can express this using the Distributive Postulate. A postulate is a statement that is true. The Distributive method is as follows:
 a (b + c) = ab + ac
 We take the 'a' and multiply it with the 'b' which gives us ab plus the 'a' multiplied by the 'c,' which gives us ac
 So, n + n ( n-3) = n + n ^ 2 - 3n = n^2 - 2n = n (n -2)
 It will not simplify to n (n-1) .
 " One item I should have asked about this expression.  If the simplified expression n(n-2) is put over denominator of 2, would this then simplify to n(n-1)  ??"
 The answer is no. Here's why. You can only cancel factors , not terms in a polynomial. Let me explain .
 2n/2 * n/2 =  n^2/2    Correct
 (2n + 5)/ 2 = n + 5    Incorrect ( You cannot cancel the 2's and get n+5)
 So, for  [n(n-2)]/2 , your answer would still be [n(n-2])/2 because there are no factors to cancel.
 You can alternately do this:
 [n(n-2)]/2 = (n^2-2n)/2= n^2/2 -2n/2 =  n^2/2 - n
 You are getting rid of the parenthesis and then putting each term over the same denominator.  The two terms are n^2 and -2n . You put each over the denominator  2. Then you can reduce each separate fraction where applicable.However, it does make things more complicated. This topic comes under
 'simplifying rational expressions."
 
 
 
 
 
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